Question

For what value of x the matrix A is singular?

$\left(\mathrm{i}\right)A=\left[\begin{array}{cc}1+x& 7\\ 3-x& 8\end{array}\right]$

$\left(\mathrm{ii}\right)A=\left[\begin{array}{ccc}x-1& 1& 1\\ 1& x-1& 1\\ 1& 1& x-1\end{array}\right]$

Answer 1

(i) Matrix A will be singular if

$\left|A\right|=0$
$$\begin{gathered} \left|A\right|=\left|\begin{array}{cc}1+x& 7\\ 3-x& 8\end{array}\right|=0 \\ \Rightarrow 8+8x-21+7x=0 \\ \Rightarrow 15x-13=0 \\ \Rightarrow 15x=13 \\ \Rightarrow x=\frac{13}{15} \end{gathered}$$

(ii) Matrix A will be singular if

$\left|A\right|=0$
$$\begin{gathered} \Rightarrow \left(x-1\right)\left[{\left(x-1\right)}^2-1\right]-1\left(x-1-1\right)+1\left[1-\left(x-1\right)\right]=0 \\ \Rightarrow \left(x-1\right)\left(x^2-2x\right)-1\left(x-2\right)+1\left(2-x\right)=0 \\ \Rightarrow x^3-2x^2-x^2+2x-x+2-x+2=0 \\ \Rightarrow x^3-3x^2+4=0 \\ \Rightarrow {\left(x-2\right)}^2\left(x+1\right)=0 \\ \Rightarrow x=2\mathrm{or}x=-1 \end{gathered}$$